Question

The functions f and g are integrable and;

Answer 1

Answer:

See explanation

Step-by-step explanation:

Given

$\int\limits^4_2 {f(x)} \, dx =4\\ \\\int\limits^7_2 {f(x)} \, dx =-5\\ \\\int\limits^7_2 {g(x)} \, dx =2$

a. $\int\limits^a_a {f(x)} \, dx =0\Rightarrow \int\limits^4_4 {f(x)} \, dx =0$

b. $\int\limits^a_b {g(x)} \, dx =-\int\limits^b_a {g(x)} \, dx\Rightarrow \int\limits^2_7 {g(x)} \, dx=-\int\limits^7_2 {g(x)} \, dx=-2$

c. $\int\limits^a_b {kg(x)} \, dx =k\int\limits^a_b {g(x)} \, dx \Rightarrow \int\limits^7_2 {4g(x)} \, dx =4\int\limits^7_2 {g(x)} \, dx =4\cdot 2=8$

d. $\int\limits^a_b {x} \, dx +\int\limits^c_a {x} \, dx =\int\limits^c_b {x} \, dx \Rightarrow \int\limits^4_2 {f(x)} \, dx +\int\limits^7_4 {f(x)} \, dx =\int\limits^7_2 {f(x)} \, dx$

Then

$\int\limits^7_4 {x} \, dx =\int\limits^7_2 {f(x)} \, dx -\int\limits^4_2 {x} \, dx =-5-4=-9$

e. $\int\limits^a_b {(gf(x)-f(x))} \, dx =\int\limits^a_b {g(x)} \, dx -\int\limits^a_b {f(x)} \, dx \Rightarrow$

$\int\limits^7_2 {(g(x)-f(x))} \, dx =\int\limits^7_2 {g(x)} \, dx -\int\limits^7_2 {f(x)} \, dx =2-(-5)=7$